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# STEP 2016 Solutions

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1. (Original post by matthewdjones)
How many marks do you think I'd get for getting all the way to the F=ma part and then stoping because I was too scared of the algebra hahaha ?
Dunno, I don't do step (I'm a bio natsci lmfao) , but judging from my instincts I think that'll be at least 8-12 marks off because that seemed to be the majority of the work (had to spot x^2=0 approximation and binomial approximation, trigonometric manipulation etc)
2. Seems no one has posted a solution to STEP III Q2, so here it is.

(i) y2 = 4ax in parametric form is: x = at2, y = 2at, where t is our parameter.

Hence t = p at P, t = q at Q, and t = r at R.

Gradient of curve at the point with parameter t is:

dy/dx = dy/dt * dt/dx = 2a / 2at = t -1.

So the gradient of the normal to the curve at the point with parameter t is -t.

Hence, the normal to the curve at Q has gradient -q, so has equation

y - 2aq = -q (x - aq2).

Since this line passes through P, we substitute y = 2ap and x = ap2. Dividing through by a, we have

2 (p - q) = -q (p2 - q2) = -q (p + q) (p - q), so 2 = -q (p + q) and hence q2 + qp + 2 = 0.

Similarly, r2 + rp + 2 = 0.

(ii) Line QR has gradient (2aq - 2ar) / (aq2 - ar2) = 2(q - r) / [ (q - r) (q + p) ] = 2 (p + q) -1.

r2 + rp + 2 = 0 = q2 + qp + 2
=> r2 - q2 = (r - q) (r + q) = p (q - r)
=> r + q = -p
=> gradient of line QR is -2 p -1
=> equation of QR is y = -2 p -1 (x - ar2) + 2ar.

Let y = 0. Then x = a (r2 + rp) = -2a.

So QR passes through ( -2a, 0). This point is clearly dependent solely on a and not on the positon of P.

(iii) Line OP has equation y = 2 p -1 x.

So, at T, where OP and QR intersect,

2 p -1 x = -2 p -1 (x - ar2) + 2ar, which implies that

x = 1/2 a (r2 + 2rp) = 1/2 a (-2) = -a.

So T lies on the line x = -a. This is clearly dependent solely on A and not on the positon of P.

Substituting x = -a into the equation of OP, we find that T has y co-ordinate -2ap -1. So the distance of the x-axis from T is the modulus of this.

Since q and r are distinct real numbers satisfying u2 + up + 2 = 0, this quadratic must have positive discriminant, i.e. p2 > 8, so the modulus of p is greater than 2 (2)1/2.

Hence the modulus of -2ap-1 is less than a / 21/2.

3. (Original post by riquix)
Seems no one has posted a solution to STEP III Q2, so here it is.

(i) y2 = 4ax in parametric form is: x = at2, y = 2at, where t is our parameter.

Hence t = p at P, t = q at Q, and t = r at R.

Gradient of curve at the point with parameter t is:

dy/dx = dy/dt * dt/dx = 2a / 2at = t -1.

So the gradient of the normal to the curve at the point with parameter t is -t.

Hence, the normal to the curve at Q has gradient -q, so has equation

y - 2aq = -q (x - aq2).

Since this line passes through P, we substitute y = 2ap and x = ap2. Dividing through by a, we have

2 (p - q) = -q (p2 - q2) = -q (p + q) (p - q), so 2 = -q (p + q) and hence q2 + qp + 2 = 0.

Similarly, r2 + rp + 2 = 0.

(ii) Line QR has gradient (2aq - 2ar) / (aq2 - ar2) = 2(q - r) / [ (q - r) (q + p) ] = 2 (p + q) -1.

r2 + rp + 2 = 0 = q2 + qp + 2
Finish it then lol.

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4. (Original post by physicsmaths)
Finish it then lol.

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lol shhh
5. (Original post by riquix)
Seems no one has posted a solution to STEP III Q2, so here it is.

(i) y2 = 4ax in parametric form is: x = at2, y = 2at, where t is our parameter.

Hence t = p at P, t = q at Q, and t = r at R.

Gradient of curve at the point with parameter t is:

dy/dx = dy/dt * dt/dx = 2a / 2at = t -1.

So the gradient of the normal to the curve at the point with parameter t is -t.

Hence, the normal to the curve at Q has gradient -q, so has equation

y - 2aq = -q (x - aq2).

Since this line passes through P, we substitute y = 2ap and x = ap2. Dividing through by a, we have

2 (p - q) = -q (p2 - q2) = -q (p + q) (p - q), so 2 = -q (p + q) and hence q2 + qp + 2 = 0.

Similarly, r2 + rp + 2 = 0.

(ii) Line QR has gradient (2aq - 2ar) / (aq2 - ar2) = 2(q - r) / [ (q - r) (q + p) ] = 2 (p + q) -1.

r2 + rp + 2 = 0 = q2 + qp + 2
=> r2 - q2 = (r - q) (r + q) = p (q - r)
=> r + q = -p
=> gradient of line QR is -2 p -1
=> equation of QR is y = -2 p -1 (x - ar2) + 2ar.

Let y = 0. Then x = a (r2 + rp) = -2a.

So QR passes through ( -2a, 0). This point is clearly dependent solely on a and not on the positon of P.

(iii) Line OP has equation y = 2 p -1 x.

So, at T, where OP and QR intersect,

2 p -1 x = -2 p -1 (x - ar2) + 2ar, which implies that

x = 1/2 a (r2 + 2rp) = 1/2 a (-2) = -a.

So T lies on the line x = -a. This is clearly dependent solely on A and not on the positon of P.

Substituting x = -a into the equation of OP, we find that T has y co-ordinate -2ap -1. So the distance of the x-axis from T is the modulus of this.

Since q and r are distinct real numbers satisfying u2 + up + 2 = 0, this quadratic must have positive discriminant, i.e. p2 > 8, so the modulus of p is greater than 2 (2)1/2.

Hence the modulus of -2ap-1 is less than a / 21/2.

Y coordinate is wrong mate for T. Should -2a/p iirc.

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6. (Original post by physicsmaths)
Y coordinate is wrong mate for T. Should -2a/p iirc.

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hence why I wrote -2ap-1 mate
7. STEP III Q1 attached.
Very nice starter question to the paper in my opinion! Will do some of the others tomorrow if I get some time

(Edit: there's a pair of limits, a pair of brackets and a power of n missing from the denominator of the first term after "integration by parts" missing but it's still very obvious what should be there - sorry, I did it quite quickly)
Attached Images

8. STEP III Q5
Solution
(i)
Note (where the latter sum is taken over the obvious range):

Hence, setting in the above:

As required.

(ii)
First s.t. .

Furthermore, ; but observe that consists of prime factors, each of which is at least , whereas each prime factor of is at most i.e. and are coprime.

It follows that and hence that , as desired.

Finally, note , as was to be deduced.

(iii)
Noting for and the result of (ii), observe that:

as required.

(iv)
Note .

Now suppose , then .

Furthermore, , as is even and therefore not prime. Hence, we also have .

It follows by strong induction that , as was to be shown.

9. (Original post by Alex_Aits)
STEP III Q1 attached.
Very nice starter question to the paper in my opinion! Will do some of the others tomorrow if I get some time

(Edit: there's a pair of limits and a pair of brackets missing but it's still very obvious what should be there - sorry, I did it quite quickly)
I did the paper earlier so for anyone who maintained their belief that the suggested expression would make the entire question easier here is a (very) slightly different solution (:P)
Attachment 556355556357556350
Attachment 556355556357

Edit: these pictures aren't coming out very clearly at all, but here is q2 and q3
Attachment 556365556367
Seems like my wifi isn't playing ball.. looks like I will have to latex the rest. Although 7,9,10 all need diagrams to do the solution justice
Attached Images

10. STEP III Q6
Latex here doesn't seem to work so i'm uploading a pdf (+latex source +geogebra but I had to add .txt to the filename otherwise TSR doesn't let me upload)

I couldn't do it all correctly. In part (iii), I forgot to consider the case b≥√(1+a²), and in part (iv) I did one of the derivations wrong. How much marks do U think I will lose?
Attached Images
11. step3q6.pdf (107.0 KB, 110 views)
12. Attached Files
13. step3q6.tex.txt (3.8 KB, 31 views)
14. step3q6.ggb.txt (6.5 KB, 18 views)
15. STEP III Question 3
Attached Images
16. STEP_2016_III_Q3.pdf (328.5 KB, 135 views)
17. STEP III Question 13
Attached Images
18. STEP_2016_III_Q13.pdf (428.7 KB, 67 views)
19. (Original post by Mathemagicien)
STEP III Question 3
Haha, I was also proud of my notation for the first part. I didn't use the partial derivative notation though, I just used D of something to denote the degree.

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20. (Original post by Farhan.Hanif93)
STEP III Q5
Solution
(i)
Note (where the latter sum is taken over the obvious range):

Hence, setting in the above:

As required.

(ii)
First s.t. .

Furthermore, ; but observe that consists of prime factors, each of which is at least , whereas each prime factor of is at most i.e. and are coprime.

It follows that and hence that , as desired.

Finally, note , as was to be deduced.

(iii)
Noting for and the result of (ii), observe that:

as required.

(iv)
Note .

Furthermore, suppose , then .

It follows by strong induction that , as was to be shown.

For the last part shouldnt you also say that p1,2m+2=p1,2m+1?
21. (Original post by Ewanclementson)
For the f(x), g(x), h(x) would an acceptable answer to the final part be the observation that it can be arranged to have (h(x)+x)+(h(1/1-x)+(1/1-x))=1 and so each of these pairs of terms are 'something' +the function of that 'something' and so it is clear that h(x)=1/2-x would be the solution as the -x part would cancel the 'something' and then the two +1/2 would add to give the one. Note: the 'something' is a function of x
No, that is not showing that it is the only solution.
22. STEP III Questions 11 and 12
Always a gift or two in the applied section
Attached Images
23. STEP_2016_III_Q11.pdf (172.3 KB, 82 views)
24. STEP_2016_III_Q12.pdf (133.1 KB, 65 views)
25. (Original post by Mathemagicien)
STEP III Question 13
Is this a valid method for part i?

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26. (Original post by fa991)
Is this a valid method for part i?
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I've only skimmed it, but it does look acceptable, yes

(And probably a better method generally, but in this case, calculating E(X^2) was very quick)
27. (Original post by gasfxekl)
For the last part shouldnt you also say that p1,2m+2=p1,2m+1?
Yes. Well spotted.
28. It was a day filled with mixed emotions. I got 3 full solutions in the first 90 mins then couldn't do much for the last 90 mins.
Score predictions anyone?
Q1 - Full
Q2 - Did the first part, said that the first part also holds for r and formed equations for OP and QR and attempted to solve for intersection points. Could not proceed through that though.
Q4 - Full except for the last part where I did not simplify and let the final answer be 2 (sech y + whatever the ans in 2nd to last part)
Q5 - Did the first part, Claimed that (2m+1)! covers Pm+1, 2m+1 and the remaining primes will cancel with m! for the 2nd part. 3rd part I stated that it holds when 2m+1 is composite and couldn't proceed for prime. 4th part I attempted induction with even n (wasn;t successful though) and stated that if it is true for even m then it is true for m+1.
Q8 - Full.
Q6 - Had 10 minutes left when I began this. Stated the values of R and gamma, and used the formula for arctanh for the conditions for A and B. I then solved for the intersection point, obtaining the result as the instructor said 'STOP WRITING'.

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